Question 1118531: The digits 0, 2, 3, 4, 5, 7, and 9 are to be formed with a three digit (without repetition) number. What is the probability that the number formed is an even number?
Found 2 solutions by stanbon, greenestamps:
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! The digits 0, 2, 3, 4, 5, 7, and 9 are to be formed with a three digit (without repetition) number. What is the probability that the number formed is an even number?
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# of 3-digit numbers:: 6*6*5 = 180
# ending in 0,2 or 4:: 6*3*5 = 90
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Ans:: P(even) = 1/2
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Cheers,
Stan H.
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Answer by greenestamps(13200)  (Show Source):
You can put this solution on YOUR website!
Total number of 3-digit numbers, since the first digit can't be 0: 6*6*5 = 180
Number of even 3-digit numbers with final digit 0: 1*6*5 = 30 (since 0 is the final digit, we don't have to worry about possibly choosing it for the first digit, which would not be allowed)
Number of even 3-digit numbers with final digit 2: 1*5*5 = 25 (only 5 choices for the first digit, since it can't be 0; then any 5 of the remaining digits for the second digit)
Number of even 3-digit numbers with final digit 4: 1*5*5 = 25 (same reason)
Total number of even 3-digit numbers: 30+25+25 = 80.
P(even) = 80/180 = 4/9
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